2.5 Q-8

Question Statement

We are given the following equations:

$$x = a \cos^3 \theta$$ $$y = b \sin^3 \theta$$

Prove that:

$$a \frac{d y}{d x} + b \tan \theta = 0$$

---

Background and Explanation

To solve this problem, we need to:

• Differentiate the given equations with respect to $\theta$ (the angle).
• Use the chain rule to connect the derivatives of $x$ and $y$ with respect to $\theta$ and eventually find $\frac{dy}{dx}$.
• The goal is to show that $a \frac{d y}{d x} + b \tan \theta = 0$.

---

Solution

Step 1: Differentiate $x = a \cos^3 \theta$ with respect to $\theta$

We are given that:

$$x = a \cos^3 \theta$$

Differentiating with respect to $\theta$:

$$\frac{d x}{d \theta} = \frac{d}{d \theta}\left(a \cos^3 \theta\right)$$

Using the chain rule:

$$\frac{d x}{d \theta} = 3a \cos^2 \theta (-\sin \theta)$$

Thus, the derivative becomes:

$$\frac{d x}{d \theta} = -3a \cos^2 \theta \sin \theta$$

To find $\frac{d \theta}{d x}$, we take the reciprocal:

$$\frac{d \theta}{d x} = -\frac{1}{3a \cos^2 \theta \sin \theta}$$

Step 2: Differentiate $y = b \sin^3 \theta$ with respect to $\theta$

Next, we are given:

$$y = b \sin^3 \theta$$

Differentiating with respect to $\theta$:

$$\frac{d y}{d \theta} = \frac{d}{d \theta}\left(b \sin^3 \theta\right)$$

Using the chain rule again:

$$\frac{d y}{d \theta} = 3b \sin^2 \theta \cdot \cos \theta$$

Step 3: Find $\frac{d y}{d x}$

Now, to find $\frac{d y}{d x}$, we apply the chain rule:

$$\frac{d y}{d x} = \frac{d y}{d \theta} \times \frac{d \theta}{d x}$$

Substitute the values we found:

$$\frac{d y}{d x} = \left(3b \sin^2 \theta \cdot \cos \theta\right) \times \left(-\frac{1}{3a \cos^2 \theta \sin \theta}\right)$$

Simplifying:

$$\frac{d y}{d x} = -\frac{b}{a} \tan \theta$$

Step 4: Final Expression

Now, multiply both sides of the equation by $a$:

$$a \frac{d y}{d x} = -b \tan \theta$$

Finally, adding $b \tan \theta$ to both sides:

$$a \frac{d y}{d x} + b \tan \theta = 0$$

Thus, we have proved the required result.

---

Key Formulas or Methods Used

• Chain Rule: To differentiate composite functions.

$$\frac{d}{d \theta} (a \cos^3 \theta) = 3a \cos^2 \theta (-\sin \theta)$$ $$\frac{d}{d \theta} (b \sin^3 \theta) = 3b \sin^2 \theta \cdot \cos \theta$$

• Reciprocal Rule: To find $\frac{d \theta}{d x}$ from $\frac{d x}{d \theta}$.

---

Summary of Steps

1. Differentiate $x = a \cos^3 \theta$ with respect to $\theta$ to find $\frac{d x}{d \theta}$.
2. Differentiate $y = b \sin^3 \theta$ with respect to $\theta$ to find $\frac{d y}{d \theta}$.
3. Use the chain rule to calculate $\frac{d y}{d x}$.
4. Multiply both sides by $a$ and simplify.
5. Conclude that $a \frac{d y}{d x} + b \tan \theta = 0$.

---